$\int^{1}_{0}(2x+1)e^{x^2+x}\,dx\, = $
Explanation: Strategy Let's first find the indefinite integral $\int(2x+1)e^{x^2+x}\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int(2x+1)e^{x^2+x}\,dx\, $, we can use U-substitution. If we let $ {u=x^2+x}$, then ${du=2x+1 \, dx}$. So we have: $\begin{aligned}\int(2x+1)e^{x^2+x}\,dx\,&=\int e^{{x^2+x}}\,\cdot {(2x+1)\,dx}\,\,\\\\\\\\ &=\int e^ u\, { du}\,\\\\\\\\ &=e^u+C\\\\\\\\ &=e^{{x^2+x}}+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{1}_{0}(2x+1)e^{x^2+x}\,dx\, &= e^{{x^2+x}}\Bigg|^1_0\\\\\\\\ &=e^2-e^0\\\\\\\\ &=e^2-1\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{1}_{0}(2x+1)e^{x^2+x}\,dx\, = e^2-1$